Problem 98 - Introductory Chemistry Atoms First

Question

At \(25^{\circ} \mathrm{C}\), the solubility of calcium oxalate, \(\mathrm{CaC}_{2} \mathrm{O}_{4}\), in water is \(6.1 \mathrm{mg} / \mathrm{L}\). (a) What are the equilibrium molar concentrations of \(\mathrm{Ca}^{2+}(a q)\) and \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) ?\) (b) Calculate \(K_{\mathrm{sp}}\) for calcium oxalate.

Step-by-Step Solution

Answer

The equilibrium molar concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) ions are both \(4.76 \times 10^{-5}\,\mathrm{mol/L}\), and the solubility product of calcium oxalate, \(K_{sp}\), is \(2.27 \times 10^{-9}\) at \(25^{\circ}\mathrm{C}\).

Step 1: Convert solubility to molar concentration for calcium oxalate

First, we need to convert the solubility of calcium oxalate from mg/L to moles/L (molar concentration). To do this, we'll divide by the molar mass of calcium oxalate: Molar mass of \(\mathrm{CaC}_{2}\mathrm{O}_{4} = 40.08\,(\mathrm{Ca}) + 24.01 \times 2\,(\mathrm{C}) + 16.00 \times 4\,(\mathrm{O}) = 128.1 \,\mathrm{g/mol}\) Now, convert the solubility into molar concentration: \(C_{\mathrm{CaC}_{2}\mathrm{O}_{4}} = \frac{6.1\,\mathrm{mg/L}}{128.1\,\mathrm{g/mol}} \times \frac{1\, \mathrm{g}}{1000\, \mathrm{mg}} = 4.76 \times 10^{-5}\,\mathrm{mol/L}\)

Step 2: Find the equilibrium molar concentrations of ions

Given that the dissolution of calcium oxalate can be represented as: \(\mathrm{CaC}_{2} \mathrm{O}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q) + \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \) Since the molar ratio of \(\mathrm{Ca}^{2+}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is \(1:1\), we'll have the same equilibrium molar concentration of \(\mathrm{Ca}^{2+}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) as the molar concentration of calcium oxalate: \(C_{\mathrm{Ca}^{2+}} = C_{\mathrm{C}_{2} \mathrm{O}_{4}^{2-}} = 4.76 \times 10^{-5}\,\mathrm{mol/L}\)

Step 3: Calculate \(K_{sp}\) for calcium oxalate

Now we can use the molar concentrations and the formula for \(K_{sp}\) to find the solubility product of calcium oxalate: \(K_{sp} = [\mathrm{Ca}^{2+}] [\mathrm{C}_{2} \mathrm{O}_{4}^{2-}]\) \(K_{sp} = (4.76 \times 10^{-5})^2 = 2.27 \times 10^{-9}\) The solubility product of calcium oxalate is \(K_{sp} = 2.27 \times 10^{-9}\) at \(25^{\circ}\mathrm{C}\).

Key Concepts

Molar Concentration

When dealing with solutions, converting quantities to molar concentration is essential. Molar concentration, also known as molarity, is defined as the number of moles of a solute per liter of solution. It tells us how many molecules or ions are present in a certain volume of solution, making it a fundamental concept in chemistry.

In the context of solubility, such as with calcium oxalate, converting solubility from milligrams per liter is crucial. To do this, you need the molar mass of the compound. In our example, the molar mass of calcium oxalate is calculated at 128.1 g/mol. By dividing the solubility given in mg/L by this molar mass (and converting units properly), we achieve the molarity. This provides the molar concentration of calcium oxalate in the solution.

Equilibrium Concentration

At equilibrium, a chemical reaction reaches a state where the concentrations of reactants and products remain constant over time. In the dissolution of calcium oxalate, the solid form and its dissociated ions achieve a balance without further changes.

For calcium oxalate, which dissociates into calcium ions \(\mathrm{Ca}^{2+} \) and oxalate ions \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \), the equilibrium concentrations of these ions can be directly observed. Since calcium oxalate dissolves in a 1:1 ratio to form each ion, their equilibrium concentrations are equal to the molar concentration of the dissolved calcium oxalate.

Calcium Oxalate Solubility

The solubility of a compound like calcium oxalate in water can tell us how much of the compound can dissolve at a given temperature. It's often expressed in units like mg/L but can be converted to molarity, which helps in calculations involving equilibrium and solubility product constants.

Solubility itself is affected by conditions such as temperature, pressure, and the presence of other ions. At 25°C, the solubility of calcium oxalate was found to be 6.1 mg/L. With the molar mass known, this can be converted to a molarity of approximately 4.76 x 10^-5 mol/L, signifying how saturated the solution is with this compound at equilibrium.

Calcium Ion Concentration

The concentration of calcium ions in a solution has a significant impact on both chemical reactivity and biological systems. In the case of calcium oxalate, calculating the concentration of \(\mathrm{Ca}^{2+} \) ions is a straightforward process, given the 1:1 molar ratio in its dissolution.

At equilibrium, the concentration of calcium ions is the same as the molarity of dissolved calcium oxalate. In our exercise, this was found to be 4.76 x 10^-5 mol/L. This concentration is not only essential for finding the solubility product \(K_{sp}\) but also useful for predicting the impact of calcium in solutions and biological scenarios.